Question

An Olympic class sprinter starts a race from rest with an acceleration of 4.8 m/s2. Take her direction of motion as the positive direction.

Answer 1

Answer:

The question is incomplete, the complete form is:

An Olympic class sprinter starts a race from rest with an acceleration of $4.8 m/s^{2}$. Take her direction of motion as the positive direction.

(a) What is her speed 2.40 s later?

(b) Sketch a graph of her position vs. time for this period.

The answer is:  

a) What is her speed 2.40 s later?

Her speed is 11.52 m/s

Explanation:

The speed can be determined by means of the kinematic equation that corresponds to a Uniformly Accelerated Rectilinear Motion.

$v = v_{0} + at$  (1)

Where $v_{0}$ is the initial velocity, a is the acceleration and t is the time.

Since she starts the race from rest her initial speed will be zero ($v_{0} = 0$)

(a) What is her speed 2.40 s later?

By means of equation 1 it is gotten:

$v = 0 m/s + (4.8 m/s^{2})(2.40s)$

$v = 11.52 m/s$

Hence, her velocity is 11.52 m/s

(b) Sketch a graph of her position vs. time for this period.

The position of that period can be known by means of the next equation:

$x = v_{0}t + \frac{1}{2}at^{2}$ (2)

However, the initial speed is equal to zero, so equation 2 can be rewritten:

$x = \frac{1}{2}at^{2}$  (3)

But the acceleration is equal to $4.8m/s^{2}$

$x = \frac{1}{2}(4.8m/s^{2})t^{2}$ (4)

Then, equation 4 can be used to sketch a graph of her position vs. time.

For t = 0 s

$x = \frac{1}{2}(4.8m/s^{2})(0)^{2}$

$x = 0$

For t = 1 s

$x = \frac{1}{2}(4.8m/s^{2})(1s)^{2}$

$x = 2.4m$

For t = 1.40 s

$x = \frac{1}{2}(4.8m/s^{2})(1.40s)^{2}$

$x = 4.70m$

For t = 2 s

$x = \frac{1}{2}(4.8m/s^{2})(2s)^{2}$

$x = 9.6m$

For t = 2.40 s

$x = \frac{1}{2}(4.8m/s^{2})(2.40s)^{2}$

$x = 13.82m$